JEE Advanced 2023 — Paper 1 — PHY-II solution

JEE Advanced2023Paper 1PHY-II

One mole of an ideal gas expands adiabatically from an initial state $(T_{A}, V_{0})$ to final state $(T_{f}, 5 V_{0})$ . Another mole of the same gas expands isothermally from a different initial state $(T_{B}, V_{0})$ to the same final state $(T_{f}, 5 V_{0})$ . The ratio of the specific heats at constant pressure and constant volume of this ideal gas is $γ$ . What is the ratio $T_{A} / T_{B}$ ?

A
$5^{γ - 1}$
B
$5^{1 - γ}$
C
$5^{γ}$
D
$5^{1 + γ}$

Solution

Isothermal process. Temperature is constant, so $T_{B} = T_{f}$ .

Adiabatic process. $T V^{γ - 1}$ is constant: $T_{A} V_{0}^{γ - 1} = T_{f} (5 V_{0})^{γ - 1}$ , so $T_{A} = T_{f} \cdot 5^{γ - 1}$ .

Ratio. $\frac{T _{A}}{T _{B}} = \frac{T _{f} \cdot 5 ^{γ - 1}}{T _{f}} = 5^{γ - 1}$ .

Answer: (A).

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